To factor the trinomial \(9x^2 - 9x - 4\) completely, we can use the method of factoring by grouping or the quadratic formula to find the roots.
We'll start with factoring:
-
Identify \(a\), \(b\), and \(c\) in the trinomial \(ax^2 + bx + c\):
- \(a = 9\)
- \(b = -9\)
- \(c = -4\)
-
Calculate the discriminant \(D = b^2 - 4ac\): \[ D = (-9)^2 - 4(9)(-4) = 81 + 144 = 225 \] Since the discriminant is a perfect square, we can factor it.
-
Find the roots using the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{9 \pm 15}{18} \]
- \(x_1 = \frac{24}{18} = \frac{4}{3}\)
- \(x_2 = \frac{-6}{18} = -\frac{1}{3}\)
-
Write the factored form using the roots: The factors corresponding to the roots are \(x - \frac{4}{3}\) and \(x + \frac{1}{3}\).
-
Adjusting the factors to eliminate the fractions: Multiply each factor by 3: \[ 3 \left( x - \frac{4}{3} \right) = 3x - 4,\quad 3 \left( x + \frac{1}{3} \right) = 3x + 1 \]
Thus, the factored form becomes: \[ (3x - 4)(3x + 1) \]
Putting this in the requested format, the answer is: \((3x - 4)(3x + 1)\)
3 answers